Calculus - optimization problem

In summary, the conversation is about finding the dimensions of an aquarium to minimize the cost of its construction. The cost formula is 2hw + 2hl + 5wl, and the volume formula is V= whl. The speaker is struggling to find the second equation needed to solve the problem, which is causing confusion in using the Lagrange multiplier method.
  • #1
Pearce_09
74
0
ok, i have an aquarium of volume V... where the cost of the base is 5 times of the sides... let h- height , w- width, l - lenght...
therefore 2hw + 2hl is the cost of the sides and 5wl is the cost of the base
therefore
total cost = 2hw + 2hl + 5wl --> min
and i want find the dimensions to minimize cost.
I know how to do problems like these I've done many, its just this one is somehow bothering me...
i need to solve for 1 variable then i have a formula with 2 variables.. then i take the partial derivatives to see if my dimensions that i find indeed minimize cost. but for some reason i can't even get past my total cost formula... help please
 
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  • #2
You know the volume formula, so that in conjunction with your cost forumula will provide you with one variable in terms of another and bingo
 
  • #3
V= whl is one equation. Can you use the "lagrange multiplier" method?
 
  • #4
i know how to use lagrange.. but i was told not to for this question
 
  • #5
I use my cost equation to solve for h and got
h = -5wl / 2w + 2l
and when i sub that into volume and partial derive but it doesn't make any sense.. I get l = w and then that implies that h = -10/8 ... not possible to have a negative number
 
  • #6
No, you don't have a cost equation. You only have a cost function that you want to minimize. You don't know what it is equal to. (It certainly isn't equal to 0!) The only equation you have is V= whl
 
  • #7
im missing an equation then, which I am having trouble finding...
 
  • #8
I just don't understand how to find this other equation
 

1. What is an optimization problem in calculus?

An optimization problem in calculus involves finding the maximum or minimum value of a function, usually with certain constraints or limitations.

2. How is calculus used to solve optimization problems?

Calculus is used to solve optimization problems by finding the derivative of the function and setting it equal to zero, then solving for the variable. This will give the critical points, which can then be plugged back into the original function to determine the maximum or minimum value.

3. What is the difference between a local maximum and a global maximum in calculus?

A local maximum is the highest point on a specific interval of a function, while a global maximum is the highest point on the entire function. In other words, a local maximum is only the highest point in a specific region, while a global maximum is the overall highest point.

4. Can calculus be used to solve real-world optimization problems?

Yes, calculus can be used to solve real-world optimization problems, such as finding the optimal production rate for a company or the optimal route for a delivery truck. Many real-world situations can be modeled using functions and optimized using calculus techniques.

5. What are some common techniques used in calculus to solve optimization problems?

Some common techniques used in calculus to solve optimization problems include the first and second derivative tests, Lagrange multipliers, and the method of substitution. These techniques involve using derivatives and setting them equal to zero to find the critical points of the function.

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